Quick Logs question - Use $\log_{24} 12$ to find $\log_{24} 2$

Question

Full Question is:

Given that $\log_{24} 12 =0.782$, find the value of $\log_{24} 2$ .

How do I / should I set this out as well, formally?

Please post correct question. – Kanwaljit Singh Jan 16 '17 at 17:09 — Kanwaljit Singh, Jan 16 '17 at 17:09

Arnaldo · Answer 1 · 2017-01-16T17:34:39.957

3

$$\log_{24}12=\log_{24}\left(\frac{24}{2}\right)=\log_{24}24-\log_{24}2=1-\log_{24}2$$

edited Jan 16 '17 at 17:34

answered Jan 16 '17 at 17:12

Arnaldo

By "hint" you mean "almost complete solution" ;-P – Fly by Night Jan 16 '17 at 17:30
3

Even easier, start with $\log_{24}2=\log_{24}\frac{24}{12}$. – David K Jan 16 '17 at 17:37
@DavidK: I think they are equivalent. The idea and the calculation. But in fact, it is another good point. – Arnaldo Jan 16 '17 at 17:40
Oh, yes, it is exactly the same idea and steps, except that you don't need to follow it with "solve for $x$ in $0.782 = 1 - x$." – David K Jan 16 '17 at 17:48
@Cicada: is it clear? – Arnaldo Jan 18 '17 at 01:28

score 1 · Answer 2 · edited Jan 16 '17 at 17:24

1

$12=\dfrac{24}{2}$

$\log_{24} 12=\log_{24}{\frac{24}{2}}=\log_{24} 24 - \log_{24} 2$

$1-\log_{24}{2}= 0.782$

$\log_{24}{2} = 0.218$

edited Jan 16 '17 at 17:24

Kanwaljit Singh

answered Jan 16 '17 at 17:17

Fawad

Correct your calculation. – Kanwaljit Singh Jan 16 '17 at 17:20
@KanwaljitSingh I can't say anything more than thanks ! – Fawad Jan 16 '17 at 17:23
1

You can now say thanks. Your welcome bro :-) – Kanwaljit Singh Jan 16 '17 at 17:24
Note that if you start by observing that $2=\frac{24}{12}$ and proceed from there, you quickly get $\log_{24}2$ isolated on one side of an equation without the risk of errors while trying to solve $1-\log_{24}{2}= 0.782$. – David K Jan 16 '17 at 17:41

2 Answers2